Wednesday, April 15, 2009

PoP QuIz

Today we had a surprise quiz. After the surprise we got handed out a worksheet for homework. Everyone make sure you do it because tommorrow Mr.K will be calling people up to show how they did their answers on the board. I will do the first part of the first question with you incase some one is confused how to go about doing them still.

The Stanford-Binet IQ scores of Canadians are normally distributed with a mean of 100 and a standard deviation of 16. According to this test, what is the probability that a randomly selected Canadian will have the following IQs?

a) an IQ less than 100

First you have to find the z-score. Now remember to get the z-score you have to subtract the mean from the number your trying to find the z-score for. Then you divide by the standard deviation. So in this case it would look like...

99-100/16=-0.0625 . Why did i not use a 100 to subtract a 100? This is because the questions asks for a score less than a 100. After finding out the z-score you want to go to shadenorm here you would put in the area you want to find. In our problem here it would look like this...shadenorm(-5,-.0625). Then on your calculator you'd get an area which would be .475082 You then multiply by 100 to make it into a percent which turns out to be 47.5%

the next scribe is Jason


Daniel said...

Good effort, but since 100 IS the mean, it has a z-score of zero. Therefor, you would use normalcdf(-5,0) which is .50 or 50%.

However, these steps are unnecessary, because a normal curve is symmetrical about the mean. If the mean is 100, 50% of the data lies below 100. Furthermore, 50% of the data lies above 100.

Daniel said...

Oh, sorry, jumped to conclusions. I see what you did about 99 ≠ 100 now. Hmm, I dunno, I guess I'll wait and see what Mr. K says, because I think that 99.99999999999999 would still technically be less than 100, so to jump down to 99 excludes quite a bit of data.... I dunno.

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